Answer
(a) At the top, the person's apparent weight is: $~~mg - m~\omega^2~R$
(b) At the bottom, the person's apparent weight is: $~~mg + m~\omega^2~R$
Work Step by Step
(a) The radial acceleration is: $~a_r = \omega^2~R$
At the top, we can find the normal force $F_N$ exerted on the person by the scale on the floor of the Ferris wheel. Note that the normal force is equal to the person's apparent weight.
$\sum F = m~a_r$
$mg-F_N = m~a_r$
$F_N = mg - m~\omega^2~R$
At the top, the person's apparent weight is: $~~mg - m~\omega^2~R$
(b) At the bottom, we can find the normal force $F_N$ exerted on the person by the scale on the floor of the Ferris wheel. Note that the normal force is equal to the person's apparent weight.
$\sum F = m~a_r$
$F_N - mg = m~a_r$
$F_N = mg + m~\omega^2~R$
At the bottom, the person's apparent weight is: $~~mg + m~\omega^2~R$.
